A card sheet divided into squares each of size 1mm² is being viewed at a distance of 9cm through a magnifying

Exercises · 9.22

1\ \text{mm}^2

(a) What is the magnification produced by the lens? How much is the area of each square in the virtual image?

(b) What is the angular magnification (magnifying power) of the lens?

(a) equal to the magnifying power in (b)? Explain.

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Concept understanding — Microscope Magnification

Microscope Magnification

A compound microscope views very small, nearby objects using two lenses in sequence: the objective (near the object) and the eyepiece (near the eye). Its total magnifying power is the product of what each lens contributes.

How the two lenses work together

The object sits just beyond the focus of the short-focal-length objective ( $f_o$ ), which forms a real, inverted, magnified image inside the tube.
That real image falls just inside the focus of the eyepiece ( $f_e$ ), which acts as a simple magnifier, producing a large virtual, magnified final image for the eye.

Because each stage magnifies, the effects multiply:

M = m_o \times m_e

The objective's magnification

m_o = \frac{v_o}{u_o} \approx \frac{L}{f_o}

where $L$ is the tube length (roughly the distance between the objective's second focal point and the eyepiece's first focal point). The object sits close to $f_o$ , so this approximation holds for a well-designed microscope.

The eyepiece's magnification

The eyepiece behaves as a simple magnifier:

Final image at the near point ( $D = 25\ \text{cm}$ , largest magnification):

m_e = 1 + \frac{D}{f_e}

Final image at infinity (relaxed eye, "normal adjustment"):

m_e = \frac{D}{f_e}

Total magnifying power

M = \frac{L}{f_o}\left(1 + \frac{D}{f_e}\right)

Important

High magnification needs short $f_o$ and $f_e$ (both sit in denominators) and a large tube length $L$ — this is why a microscope objective is always a very short-focus lens.

Worked example

Objective $f_o = 1.0\ \text{cm}$ , eyepiece $f_e = 2.5\ \text{cm}$ , tube length $L = 20\ \text{cm}$ , near point $D = 25\ \text{cm}$ . Find $M$ with the final image at the near point.

m_o = \frac{20}{1.0} = 20, \qquad m_e = 1 + \frac{25}{2.5} = 11

$M = 20 \times 11 = 220$ …

Why this formula?

Microscope Magnification: Why the Formula Holds

Let's build the understanding from first principles — not just memorizing formulas, but seeing why they work.

1. What Does "Magnification" Mean in a Microscope?

A microscope creates a larger apparent image of a tiny object. The total magnification is the product of two stages:

Objective lens — creates a real, enlarged, inverted image of the specimen.
Eyepiece (ocular) — acts as a simple magnifier to view that real image.

So:

Total magnification = (magnification by objective) × (magnification by eyepiece)

2. The Key Formula

For a compound microscope in normal adjustment (final image at infinity, relaxed eye):

M = M_o \times M_e = \left( \frac{L}{f_o} \right) \times \left( \frac{D}{f_e} \right)

Where:

$f_o$ = focal length of objective
$f_e$ = focal length of eyepiece
$L$ = tube length (distance between second focal point of objective and first focal point of eyepiece)
$D$ = near point distance of the eye (usually 25 cm)

3. Derivation of Objective Magnification $M_o$

Step 1: How the objective works

The objective lens forms a real, inverted, enlarged image of the specimen. The specimen is placed just outside its focal point ( $f_o$ ).

Step 2: Using the lens formula

For a thin lens:

\frac{1}{v_o} - \frac{1}{u_o} = \frac{1}{f_o}

(Using sign convention: $u_o$ is negative, $v_o$ is positive)

Step 3: The tube length approximation

In a standard microscope, the specimen is placed very close to $f_o$ , so:

$u_o \approx -f_o$ (object just beyond focal point)
The image is formed at the first focal point of the eyepiece, which is at a distance $L$ from the second focal point of the objective.

Thus:

v_o \approx f_o + L

Step 4: Magnification formula

Lateral magnification by objective:

M_o = \frac{v_o}{|u_o|} \approx \frac{f_o + L}{f_o} = 1 + \frac{L}{f_o}

Since $L \gg f_o$ in practice, $1$ is negligible:

\boxed{M_o \approx \frac{L}{f_o}}

Why this makes sense: A shorter $f_o$ means the objective is more "powerful" — it bends light more sharply, creating a larger image at the fixed tube length.

4. Derivation of Eyepiece Magnification $M_e$

Step 1: The eyepiece as a simple magnifier

The eyepiece takes the real image from the objective and acts like a magnifying glass. For relaxed eye (final image at infinity), the real image must be placed at the focal point of the eyepiece.

Step 2: Angular magnification

Angular magnification is defined as:

M_e = \frac{\text{angle subtended by image at eye}}{\text{angle subtended by object at near point}}

For a simple magnifier with image at infinity:

M_e = \frac{D}{f_e}

Where $D = 25$ cm (standard near point). …

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The object sits exactly at the focus ( $u = f = 9\ \text{cm}$ ), so the image forms at infinity: the linear magnification $m = v/u$ is infinite and the "area" of a square in the virtual image is likewise undefined (infinite). The angular magnification (magnifying power) is $M = D/f = 25/9 \approx 2.8$ . The two are not equal, because they measure different things — actual size versus angular size.

(a) Linear magnification and image area

The lens has $f = +9\ \text{cm}$ and the object is at $u = -9\ \text{cm}$ . From the lens equation:

\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{9} - \frac{1}{9} = 0 \;\Rightarrow\; v = \infty.

The image is formed at infinity (emergent rays are parallel). The linear magnification is

m = \frac{v}{u} \to \infty,

so the size — and hence the area — of each square in the virtual image is not finite. An image at infinity has no well-defined linear size; its area is effectively infinite. In practice one describes such an image by the angle it subtends, not by a length or area.

(b) Angular magnification (magnifying power)

With the image at infinity and the lens close to the eye, the angle subtended by the image equals the angle the object subtends at the lens, $\theta' \approx h/f$ , while unaided the object would be held at the near point, subtending $\theta \approx h/D$ . Hence

M = \frac{\theta'}{\theta} = \frac{D}{f} = \frac{25}{9} \approx 2.8.

(c) Are they equal? …

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Method: Lens Formula & Magnification Approach

This problem uses two distinct concepts — linear magnification (image size) and angular magnification (magnifying power). We solve step-by-step.

(a) Linear magnification and image area

Step 1: Identify given data

Focal length, $f = 9\ \text{cm}$
Object distance, $u = -9\ \text{cm}$ (negative by sign convention)
Object is at the focus of the lens.

Step 2: Apply lens formula

\frac{1}{v} - \frac{1}{u} = \frac{1}{f}

Substitute $u = -9\ \text{cm}$ , $f = 9\ \text{cm}$ :

\frac{1}{v} - \frac{1}{-9} = \frac{1}{9}

\frac{1}{v} + \frac{1}{9} = \frac{1}{9}

\frac{1}{v} = 0 \quad \Rightarrow \quad v = \infty

Step 3: Linear magnification

m = \frac{v}{u} = \frac{\infty}{-9} = \infty

So the linear magnification is infinite — the image is formed at infinity.

Step 4: Area of each square in the virtual image

Since the image is at infinity, the area is not defined in the usual sense.
However, the angular size of each square is what matters (see part b).
Answer: The concept of area in the image is meaningless here; the image is at infinity.

(b) Angular magnification (magnifying power)

Step 1: Formula for magnifying power when image is at infinity

M = \frac{D}{f}

where $D = 25\ \text{cm}$ (least distance of distinct vision for normal eye).

Step 2: Substitute values

M = \frac{25}{9} \approx 2.78

Step 3: Interpretation

The angular magnification is 2.78 — the angle subtended by the image is 2.78 times that of the object placed at $D$ .

(c) Comparison and explanation

Quantity	Value	Meaning
Linear magnification (a)	$\infty$	Image at infinity, size undefined

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🔍 Mistake 1: Confusing Linear Magnification with Angular Magnification

The error:

Students often think the magnification in part (a) and part (b) are the same quantity, so they give the same answer for both.

Why it’s wrong:

Linear magnification ( $m$ ) is the ratio of image height to object height.
Angular magnification ( $M$ ) is the ratio of the angle subtended by the image at the eye to the angle subtended by the object at the near point (usually $D = 25\ \text{cm}$ ).

How to avoid:

Always check what the question asks:

If it says “magnification produced by the lens” → use $m = v/u$ .
If it says “angular magnification” or “magnifying power” → use $M = D/f$ (for relaxed eye) or $M = 1 + D/f$ (for image at near point).

🔍 Mistake 2: Using the Wrong Formula for Angular Magnification

The error:

Using $M = \frac{D}{f}$ when the image is at the near point, or using $M = 1 + \frac{D}{f}$ when the eye is relaxed.

Why it’s wrong:

For a relaxed eye (image at infinity): $M = \frac{D}{f}$ .
For image at the near point (least distance of distinct vision, $D = 25\ \text{cm}$ ): $M = 1 + \frac{D}{f}$ .

How to avoid:

Read the problem carefully. Here, the lens is held close to the eye and the object is at $9\ \text{cm}$ (focal length also $9\ \text{cm}$ ). That means the image is at infinity → relaxed eye → use $M = D/f$ .

🔍 Mistake 3: Forgetting to Convert Units

The error:

Using $f = 9\ \text{cm}$ and $D = 25\ \text{cm}$ without converting to meters, or mixing cm and mm.

Why it’s wrong:

Formulas like $M = D/f$ are unitless ratios, so units must match. If $f$ is in cm, $D$ must also be in cm.

How to avoid:

Always write units beside every number. Convert everything to the same unit before plugging in.

🔍 Mistake 4: Calculating Area Magnification Incorrectly

The error:

Thinking area magnification = linear magnification $m$ .

Why it’s wrong:

Area scales as the square of linear magnification.

If $m$ is the linear magnification, then:

\text{Area magnification} = m^2

How to avoid:

First find $m$ using $m = v/u$ .
Then multiply the original area by $m^2$ .

🔍 Mistake 5: Assuming $v$ is Always Positive

The error:

Using $v$ as positive without checking sign conventions.

Why it’s wrong:

For a virtual image (as in a simple microscope), $v$ is negative in the Cartesian sign convention (object on left, image on same side as object).

How to avoid:

Use the lens formula with proper signs:

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL4 marks
$Q. A person with a normal near point (25 cm) using a compound microscope with objective of focal length 8.0 mm and an eyepiece of focal length 2.5 cm can bring an object placed at 9.0 mm from the objective in sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope.$
›Reveal solutionSolution

First find where the objective lens forms its (real, magnified) intermediate image using the lens formula; then require the eyepiece to form the final image at the near point (25 cm) for maximum magnification, which fixes the eyepiece's object distance; the lens separation is the sum of the two image/object distances, and total magnification is the product of the two individual magnifications.

Given: near point D = 25 cm, objective focal length f_o = 8.0 mm = 0.8 cm, object distance from objective u_o = 9.0 mm = 0.9 cm (object just outside focal length, as required for a microscope objective), eyepiece focal length f_e = 2.5 cm.

Step 1 - Image formed by objective: Using the thin lens formula (sign convention: distances measured from the lens, object distance taken negative):

u_o = -0.9 cm, f_o = +0.8 cm (convex lens)

1/v_o - 1/u_o = 1/f_o

1/v_o = 1/f_o + 1/u_o = 1/0.8 + 1/(-0.9) = 1.25 - 1.111 = 0.139

v_o = 1/0.139 = 7.2 cm

So the objective forms a real, inverted, magnified image at 7.2 cm on the far side of the objective.

Step 2 - Eyepiece positioned for image at near point (maximum magnification, image at D): This intermediate image (from the objective) acts as the object for the eyepiece. For maximum magnifying power, the eyepiece should form its final (virtual) image at the near point, i.e. v_e = -25 cm (virtual image, same side as the object, by sign convention).

1/v_e - 1/u_e = 1/f_e

1/(-25) - 1/u_e = 1/2.5

-0.04 - 1/u_e = 0.4

-1/u_e = 0.44

u_e = -1/0.44 = -2.27 cm

So the intermediate image must be 2.27 cm in front of the eyepiece for the eyepiece to form the final image at the near point.

…

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GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL4 marks
$Q. Draw the ray diagram for the formation of image by a compound microscope and obtained the formula for magnification.$
›Reveal solutionSolution
Figure — The stem instructs 'Draw the ray diagram for the formation of image by a compound microscope'; the catalog fig

A compound microscope's objective forms a real, magnified image which then acts as the object for the eyepiece, which forms a further-magnified virtual image; total magnification is the product of the two individual magnifications.

Ray diagram (described): The object AB is placed just beyond the focus Fo of the objective lens (between fo and 2fo). The objective forms a real, inverted, magnified image A'B' beyond 2Fo on the other side. This intermediate image A'B' falls within the focal length of the eyepiece (between the eyepiece and its focus Fe). The eyepiece then acts as a simple magnifier on this intermediate image, producing a final virtual, further-magnified, and (relative to the object) inverted image A''B'' — formed at the near point D for maximum magnification, or at infinity for normal (relaxed-eye) adjustment. Both lenses are convex, mounted coaxially at opposite ends of a tube, with the objective having a small aperture and focal length, and the eyepiece having a larger aperture.

…

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GSEB Higher Secondary Certificate (HSC) Examination 2023Set ANNUAL4 marks
$Q. A compound microscope consists of an objective lens of focal length 2.0 cm and an eye piece of focal length 6.25 cm separated by a distance 15 cm. How far from the objective should an object be placed in order to obtain the final image at$
a) the least distance of distinct vision (25 cm) and
b) at infinity? What is the magnifying power of microscope in each case?
›Reveal solutionSolution
Work back from the eyepiece: (a) for image at 25 cm, u_e = -5 cm, objective forms image at 10 cm, u_o = -2.5 cm, M = 20; (b) for image at infinity, u_e = -6.25 cm, objective image at 8.75 cm, u_o = -2.59 cm, M = 13.5.

Data: f_o = 2.0 cm, f_e = 6.25 cm, tube separation L = 15 cm, D = 25 cm.

Final image at least distance of distinct vision (25 cm): Eyepiece: image virtual at v_e = -25 cm. 1/v_e - 1/u_e = 1/f_e: 1/(-25) - 1/u_e = 1/6.25 => -1/u_e = 1/6.25 + 1/25 = 5/25 = 1/5 => u_e = -5 cm. So the objective's image is 5 cm in front of the eyepiece; distance from objective = L - 5 = 10 cm => v_o = +10 cm. Objective: 1/v_o - 1/u_o = 1/f_o: 1/10 - 1/u_o = 1/2 => -1/u_o = 1/2 - 1/10 = 2/5 => u_o = -2.5 cm. Magnifying power: M = m_o (1 + D/f_e) = (v_o/|u_o|)(1 + D/f_e) = (10/2.5)(1 + 25/6.25) = 4 x 5 = 20.

Final image at infinity: …
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Q.A card sheet divided into squares each of size 1 mm21\ \text{mm}^21 mm2 is being viewed at a distance of 9 cm9\ \text{cm}9 cm through a magnifying glass (a converging lens of focal length 9 cm9\ \text{cm}9 cm) held close to the eye.

(a) Linear magnification and image area

(b) Angular magnification (magnifying power)

(c) Are they equal? …

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Q.A card sheet divided into squares each of size $1\ \text{mm}^2$ is being viewed at a distance of $9\ \text{cm}$ through a magnifying glass (a converging lens of focal length $9\ \text{cm}$ ) held close to the eye.

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